Proof. The invariant factors are integers \( n_i \geq 2\) with \( \cdots n_3 | n_2 | n_1, \) where the prime factors of \( 100 \) (in this case) divide \( n_1 \) and \( n_1 \cdot n_2 \cdot n_3 \cdots = 100. \) Considering that \( 100 = 2^2 \cdot 5^2, \) there are four candidates for \( n_1 \), namely, \( n_1 = 2 \cdot 5, \; 2^2 \cdot 5, \; 2 \cdot 5^2, \; 2^2 \cdot 5^2. \) Note that the \( \mathbb{Z} \)-module factors in accordance with its invariant factors, so the first three don't work. Were \( n_1 = 2 \cdot 5 = 10 \), for example, then \( \mathbb{Z}_{100} \cong \mathbb{Z}_{10} \oplus N \) as \( \mathbb{Z} \)-modules (ie., groups), and \( |N| = 10. \) But the right side has no elements of order \( 100 \) like \( \mathbb{Z}_{100} \) does. (Note the notation \( \mathbb{Z}_{n} \) for \( \mathbb{Z} / n \mathbb{Z}.) \) Therefore \( n_1 = 2^2 \cdot 5^2 = 100 \) is the only invariant factor of \( \mathbb{Z}_{100}, \) which therefore is its own invariant factor decomposition.

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